具有变量列的Django查询过滤器 [英] Django query filter with variable column

问题描述

我试图使用

info=members.filter(name__contains=search_string)

我遇到的问题是我不知道用户想要提前搜索哪个字段，所以我需要替换'name 'with a variable as as

The problem I have is I do not know which field the user wants to search ahead of time so I need to substitute 'name' with a variable as in

variable_column = 'name'
search_type = 'contains'
filter = variable_column + '__' + search_type
info=members.filter(filter=search_string)

如何做？

Rich

推荐答案

几乎在那里。 。

members.filter（** {'string__contains'：'search_string'}）

要了解它在做什么，google在左边：）
了解Python中的kwargs

To understand what it's doing, google around : ) Understanding kwargs in Python

** 将字典键/值对扩展为keywo rd参数 - 值对。

** expands dictionary key/value pairs to keyword argument - value pairs.

要使您的示例适应解决方案：

To adapt your example to the solution:

variable_column = 'name'
search_type = 'contains'
filter = variable_column + '__' + search_type
info=members.filter(**{ filter: search_string })

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